Find the matrix $A\in \mathbb{R}^{3\times 3}$

Let $$A\in \mathbb{R}^{3\times 3}$$ such that $$\mathrm{rank} \,A=2$$ and the elements over the main diagonal are equal. Find the matrix $$A$$ if $$\lambda=2$$ is an eigenvalue and $$x_1=[1,2,-1]^T$$ and $$x_2=[3,0,1]^T$$ are the eigenvectors corresponding to $$\lambda=2$$.

I know I should write my attempt what I tried to do but to be honest I don't even know where to begin.

I tried writing the matrix with ordinary entries so that I can find the characteristic polynomial but I didn't get anywhere.

I would really appreciate some help. Please don't downvote for not writing my work.

• If the matrix is of full rank then $A = Q \Lambda Q^{-1}$ where $\Lambda$ is a diagonal matrix of eigenvalues, and $Q$ is the matrix of column eigenvectors that correspond to each eigenvalue. This doesn't help in your situation, but would be a useful shortcut for other users who might stumble across this question in the future. – OmnipotentEntity Mar 29 at 1:01

Since the matrix is not full rank (that is $$\operatorname{rk}(A)\neq 3$$) we have that one eigenvalue of $$A$$ is $$0.$$ Let the elements on the diagonal be equal to say $$x.$$ Then $$\operatorname{Trace}{A} = \lambda_1+\lambda_2+\lambda_3= 2+2 + 0 = 3x$$ and so $$x= 4/3.$$
Now we know that $$Ax = \begin{bmatrix} 4/3 & a & b \\ c & 4/3 & d \\ e & f & 4/3 \\ \end{bmatrix} \begin{bmatrix} 1\\ 2\\ -1\\ \end{bmatrix} = 2\begin{bmatrix} 1\\ 2\\ -1\\ \end{bmatrix}$$ and so $$4/3 + 2a -b = 2$$ $$c + 8/3 - d = 4$$ $$e+ 2f -4/3= -2$$ Similarily you can write a system for the eigenvector. This gives you $$6$$ equations for $$6$$ variables and so you can obtain a solution for this system (hopefully the problem is well posed).